Solve for $x$ : $2x^2 + 20x + 42 = 0$
Answer: Dividing both sides by $2$ gives: $ x^2 + {10}x + {21} = 0 $ The coefficient on the $x$ term is $10$ and the constant term is $21$ , so we need to find two numbers that add up to $10$ and multiply to $21$ The two numbers $7$ and $3$ satisfy both conditions: $ {7} + {3} = {10} $ $ {7} \times {3} = {21} $ $(x + {7}) (x + {3}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 7) (x + 3) = 0$ $x + 7 = 0$ or $x + 3 = 0$ Thus, $x = -7$ and $x = -3$ are the solutions.